An empty cone is of radius 3 cm and height 12 cm. Ice-cream is filled in it so that lower part of the cone, which is 1/6th of volume of the cone, is unfilled but hemisphere is formed on the top. Find volume of the of ice-cream.

Question

Q) An empty cone is of radius 3 cm and height 12 cm. Ice-cream is filled in it so that lower part of the cone, which is $\frac{1}{6}$ th of volume of the cone, is unfilled but hemisphere is formed on the top. Find volume of the of ice-cream.

Sapling Academy · Accepted Answer

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Volume of the cone = $\frac{1}{3}\pi$ r2h = $\frac{1}{3} \pi$ x (3)2 x 12 = 36 $\pi$

When Ice cream is filled in this cone, its  $\frac{1}{6}$th portion is unfilled and  $\frac{5}{6}$th gets filled.

Hence, volume of this  $\frac{5}{6}$th cone = $\frac{5}{6}$ x 36 $\pi$  = 30 $\pi$ ………. (i)

Volume of Icecream’s hemispeherical shape on top of the cone

=        $\frac{2}{3} \pi$ r3     =        $\frac{2}{3} \pi$ x (3)3  = 18 $\pi$ ..……… (ii)

From equations (i) and (ii), we get,

Total Volume of Icecream = 30 $\pi$ + 18 $\pi$ = 48 $\pi$  = 48 x $\frac{22}{7}$

=          150.72 cm3

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